## Calculus with Applications (10th Edition)

There is a discontinuity when $a=3$ or $a=-2$
We are given $j(x)=\ln|\frac{x+2}{x-3}|$ There is a discontinuity when $a=3$ or $a=-2$ $\lim\limits_{x \to -2}j(x)=-\infty$ $\lim\limits_{x \to 3}j(x)=\infty$ The limit $\lim\limits_{x \to 3}j(x)$ and $\lim\limits_{x \to -2}j(x)$ do not exist.